Question

The minimum value of $a\sec x+bco\sec x$, $0<a<b$, $0<x<\pi /2$ is-

• A. $a+b$
• B. $a^{2/3}+b^{2/3}$
• C. ${(a^{2/3}+b^{2/3})}^{3/2}$
• D. None of these

Answer 1

The correct option is C ${(a^{2/3}+b^{2/3})}^{3/2}$

Let $y=a\sec x+b\text{cosec}x$
Therefore, $\frac{dy}{dx}=a\sec x\tan x-b\text{cosec}x\cot x$
When $\frac{dy}{dx}=0$,
Then $a\sec x\tan x=b\text{cosec}x\cot x$

$\frac{a\sin x}{{\cos }^{2}x}=\frac{b\cos x}{{\sin }^{2}x}$

$a{\sin }^{3}x=b{\cos }^{3}x$

${\tan }^{3}x=\frac{b}{a}$

$\tan x={\left(\frac{b}{a}\right)}^{1/3}$

$x={\tan }^{-1}{\left(\frac{b}{a}\right)}^{1/3}$

$⟹\sec x=\frac{\sqrt{a^{2/3}+b^{2/3}}}{a^{1/3}},cosecx=\frac{\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}}$

Minimum value
$y=a\times \sec x+b\text{cosec}x$

$y=\frac{a\times \sqrt{a^{2/3}+b^{2/3}}}{a^{1/3}}+\frac{b\times \sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}}$
$y={(a^{2/3}+b^{2/3})}^{3/2}$